At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a g

1. The total moles of the solution is 0.3079193 mol. 2. The mole fraction for gold is 0.2473212, and for silver, it is 0.7526787. 3. The molar entropy of mixing for gold is 2.87285 j/K, while for silver, it is 1.77804 j/K. 4. The total entropy of mixing sums to 4.65089 j/K. 5. Molar free energy amounts to -2325.445 kJ. 6. Chemical potential for silver is -1750.31129 j/mol and for gold, it is -575.13185 j/mol. To elaborate: (1) The molar mass of silver stands at 107.8682 g/mol, and gold's at 196.96657 g/mol. Hence, calculating moles leads to mass/molar mass for silver: 25 g/107.8682 g/mol = 0.2317643 mol and for gold: 15 g/196.96657 g/mol = 0.076155 mol, resulting in a total of 0.30791193 mol. (2) For the mole fractions, silver's fraction is 0.2317643/0.3079193 = 0.7526787, and for gold, it's 0.076155/0.3079193 = 0.2473212. (3) To find molar entropy mixing ∆Sm, we use the formula ∆Sm = -R * Xi * ln(Xi) where R = 8.3144598. For silver, substituting gives us 1.77804 j/K, while for gold, we get 2.87285 j/K. (4) Overall entropy of mixing totals 4.65089 j/K thus calculated. (5) The Gibbs free energy at 500 °C can be derived through G = H - TS, accounting to H = 0 (as T is 500 + 273 = 773 K and S is 4.65089), resulting in G equating to -3595.138 kJ. (6) The chemical potentials calculated derive from multiplying the Gibbs free energy by their mole fractions.

Answer:NH₃/NH₄Cl

Explanation:

The pH of a buffer can be determined using Henderson-Hasselbalch's equation.

When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.

Below are the buffers and their corresponding pKa values:

• CH₃COONa/CH3COOH (pKa = 4.74)

• NH₃/NH₄Cl (pKa = 9.25)

• NaOCl/HOCl (pKa = 7.49)

• NaNO₂/HNO₂ (pKa = 3.35)

• NaCl/HCl Not a buffer

Thus, the ideal buffer is NH₃/NH₄Cl.

Answer:

The designation of 70% (vol/vol) indicates

that it contains 70% (vol/vol), meaning 70 ml of isopropanol is included in 100 ml of rubbing alcohol solution.

If it were 200 ml, then naturally, it would contain 70*2 = 140 ml of isopropanol required.

Response:

1.98 M

Clarification:

Provided data

• Starting volume (V₁): 93.2 mL

• Starting concentration (C₁): 2.03 M

• Water volume added: 3.92 L

Step 1: Convert V₁ to liters

Using the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)

The final volume is the total of the initial volume and the added water volume.

Step 3: Calculate the final concentration (C₂)

Utilizing the dilution rule.